LpL_p space

Last updated: 2026-04-12

This article explores the space of functions whose integrals have a finite p-th norm. Such spaces are fundamental in many areas of mathematics, particularly in analysis, and they play crucial roles in various applications across physics, engineering, and probability.
Consider (X,X,μ)(X, \mathcal{X}, \mu), a measure space equipped with a measure μ\mu with values in [0,∞][0,\infty]) and 1≤p<∞1 \leq p < \infty. Denote by Lp(X,μ)L_p(X, \mu) the set of all μ\mu-measurable functions f:X→Rf: X \rightarrow \mathbb{R} for which ∣f(⋅)∣p|f(\cdot)|^p is a μ\mu-integrable function:
∥f∥p:=∥f∥Lp(X,μ)=(∫X∣f(x)∣p dμ(x))1/p<∞,f∈Lp(μ).\begin{equation*}\|f\|_p := \|f\|_{L_p(X, \mu)} = \left(\int_X |f(x)|^p \, \text{d}\mu(x) \right)^{1/p} < \infty, \quad f \in L_p(\mu).\end{equation*}
When considering the Lebesgue measure μ=λ\mu = \lambda on Rn\mathbb{R}^n or on a set Ω⊂Rn\Omega \subset \mathbb{R}^n the symbol Lp(Ω)L_p(\Omega) are used without specifying the Lebesgue measure λ\lambda. In place of Lp([a,b])L_p([a, b]) and Lp([a,+∞))L_p([a, +\infty)) it is customary to write Lp[a,b]L_p[a, b] and Lp[a,+∞)L_p[a, +\infty). For instance, ff lays inside Lp[a,b]L_p[a, b] if
∥f∥Lp[a,b]=(∫ab∣f(x)∣p dx)1/p<∞.\begin{equation*}\|f\|_{L_p[a, b]} = \left(\int_a^b |f(x)|^p \, \text{d} x \right)^{1/p} < \infty.\end{equation*}
Commonly used pp values include p=1,2,∞p=1, 2, \infty, with L2(X,μ)L_2(X, \mu) being particularly important due to its Hilbert space structure.
A function ff is said to be essential bounded if there is a constant K<∞K < \infty such that ∣f(x)∣≤K|f(x)| \leq K μ−\mu-a.e. on XX. The greatest lower bound of such constants KK is called essential supremum of ∣f∣|f| on XX, and is denoted by ess sup⁡x∈X∣u(x)∣\operatorname{ess\,sup}_{x \in X} |u(x)|. The vector space of all function ff that are essentially bounded on XX is denote by L∞(X,μ)L^{\infty}(X, \mu) with the norm
∥f∥∞:=ess sup⁡x∈X∣f(x)∣=inf⁡{M>0:∣f(x)∣≤Mμ−a.e.}.\begin{equation*}\| f \|_{\infty} := \operatorname{ess\,sup}_{x \in X} |f(x)| = \inf \{ M > 0: |f(x)| \leq M \quad \mu-\text{a.e.} \}.\end{equation*}

Motivation

Motivation behind LpL_p spaces

Physical Origins

The initial motivation came from classical physics:
  • L2L_2 emerged naturally in studying energy of signals
  • L1L_1 appeared in total variation and mass calculations

Functional Analysis Development

The development of LpL_p spaces was driven by:
  • Search for complete spaces of functions
  • Desire to understand convergence of Fourier series

Hölder and Minkowski inequalities

This section covers two most important inequality for Lp(X,μ)L_p(X, \mu) spaces: Hölder's and Minkowski's inequalities.

Hölder inequality

Theorem (Hölder inequality)
Suppose that 1≤p≤∞1 \leq p \leq \infty and q=pp−1q = \frac{p}{p-1}. If f∈Lp(X,μ)f \in L_p(X, \mu) and g∈Lq(X,μ)g \in L^q(X, \mu), then fg∈L1(X,μ)fg \in L^1(X, \mu). Moreover,
∫X∣f(x)g(x)∣ dμ(x)≤(∫X∣f(y)∣p dμ(y))1/p⋅(∫X∣g(z)∣q dμ(z))1/q.\begin{equation*}\int_X |f(x)g(x)| \, d\mu(x) \leq \left( \int_X |f(y)|^p \, d\mu(y) \right)^{1/p} \cdot \left( \int_X |g(z)|^q \, d\mu(z) \right)^{1/q}.\end{equation*}
or in the norm form
∥fg∥L1(X,μ)≤∥f∥Lp(X,μ)⋅∥g∥Lq(X,μ)\begin{equation*}\|fg\|_{L^1(X, \mu)} \leq \|f\|_{L_p(X, \mu)} \cdot \|g\|_{L^q(X, \mu)}\end{equation*}
\begin{proof}
\quad Proof. Let a,b>0a, b > 0 and let A=ln⁡(ap)A = \ln(a^p) and B=ln⁡(bq)B = \ln(b^{q}). Since the exponential function is strictly convex,
exp⁡(Ap+Bq)≤exp⁡Ap+exp⁡Bq,\begin{equation*}\exp \left( \frac{A}{p} + \frac{B}{q} \right) \leq \frac{\exp A}{p} + \frac{\exp B}{q} ,\end{equation*}
with equality only if A=BA = B. Hence
exp⁡(ln⁡(ap)p+ln⁡(bp)q)=ab≤app+bqq.\begin{equation*}\exp \left( \frac{\ln(a^p)}{p} + \frac{\ln(b^p)}{q} \right) = ab \leq \frac{a^p}{p} + \frac{b^{q}}{q}.\end{equation*}
The function fgfg is defined a.e. and measurable. If we set a=∣f(x)∣/∥f∥pa = |f(x)| / \|f\|_p and b=∣g(x)∣/∥g∥pb = |g(x)| / \|g\|_p then by previous inequality we get
∣f(x)g(x)∣∥f∥p∥g∥q≤1p∣f(x)∣p∥f∥pp+1q∣g(x)∣q∥g∥qq.\begin{equation*}\frac{|f(x)g(x)|}{\|f\|_p\|g\|_q} \leq \frac{1}{p} \frac{|f(x)|^p}{\|f\|_p^p} + \frac{1}{q} \frac{|g(x)|^q}{\|g\|_q^q}.\end{equation*}
If we integrate both sides then the right-hand integral integral is 1, so the left-hand side is also integrable and its integral does not exceed 1, which is exactly the Hölder inequality. □\Box
\end{proof}
\begin{corollary}
Under the hypotheses of the Hölder inequality,
∫Xfg dμ≤(∫X∣f∣p dμ)1/p(∫X∣g∣q dμ)1/q.\begin{equation*}\int_X fg \, d\mu \leq \left( \int_X |f|^p \, d\mu \right)^{1/p} \left( \int_X |g|^q \, d\mu \right)^{1/q}.\end{equation*}
\end{corollary}
\begin{proof}
\quad Proof. It comes from the following obvious inequality
∫Xfg dμ≤∫X∣fg∣ dμ□\begin{equation*}\int_X fg \, \text{d}\mu \leq \int_X | fg | \, \text{d} \mu \quad \Box\end{equation*}
\end{proof}
An immediate corollary of Hölder’s inequality is the following Cauchy–Schwarz inequality
\begin{corollary}[Cauchy–Schwarz inequality]
If f∈L2(μ)f \in L^2(\mu), g∈L2(μ)g \in L^2(\mu). Then fg∈L1(μ)fg \in L^1(\mu) and
∫Xf(x)g(x) dμ(x)≤(∫X∣f(y)∣2 dμ(y))1/2(∫X∣g(z)∣2 dμ(z))1/2.\begin{equation*}\int_X f(x)g(x) \, d\mu(x) \leq \left( \int_X |f(y)|^2 \, d\mu(y) \right)^{1/2} \left( \int_X |g(z)|^2 \, d\mu(z) \right)^{1/2}.\end{equation*}
\end{corollary}

Minkowski’s inequality

To establish the Minkowski inequalities, we first introduce the following simple lemma.
Lemma
If 1≤p<∞1 \leq p<\infty and a,b≥0a, b \geq 0, then
(a+b)p≤2p−1(ap+bp).\begin{equation*}(a+b)^p \leq 2^{p-1}\left(a^p+b^p\right) .\end{equation*}
\begin{proof}
\quad Proof. If p=1p=1, then equality holds. For p>1p>1, the function tpt^p is convex on [0,∞)[0, \infty); that is, its graph lies below the chord line joining the points (a,ap)\left(a, a^p\right) and (b,bp)\left(b, b^p\right). Thus
(a+b2)p≤ap+bp2,\begin{equation*}\left(\frac{a+b}{2}\right)^p \leq \frac{a^p+b^p}{2},\end{equation*}
from which the statement of lemma follows at once. □\Box
\end{proof}
Theorem (Minkowski’s inequality)
Let 1≤p≤∞ 1 \leq p \leq \infty and let f,g∈Lp(X,μ)f, g \in L_p(X, \mu). Then f+g∈Lp(μ)f + g \in L_p(\mu) and
(∫X∣f(x)+g(x)∣p dμ(x))1/p≤(∫X∣f(x)∣p dμ(x))1/p+(∫X∣g(x)∣p dμ(x))1/p\begin{equation*}\left( \int_X |f(x) + g(x)|^p \, \text{d} \mu(x) \right)^{1/p} \leq \left( \int_X |f(x)|^p \, \text{d} \mu(x) \right)^{1/p} + \left( \int_X |g(x)|^p \, \text{d} \mu(x) \right)^{1/p}\end{equation*}
or in the norm form
∥f+g∥Lp(X,μ)≤∥f∥Lp(X,μ)+∥g∥Lp(X,μ).\begin{equation*}\|f + g\|_{L_p(X, \mu)} \leq \|f\|_{L_p(X, \mu)} + \|g\|_{L_p(X, \mu)} .\end{equation*}
\begin{proof} \quad Proof. Consider the simple inequality
∣f+g∣p=∣f+g∣⋅∣f+g∣p−1≤∣f∣⋅∣f+g∣p−1+∣g∣⋅∣f+g∣p−1,\begin{equation*}|f + g|^p = |f+g| \cdot |f+g|^{p-1} \leq |f|\cdot|f+g|^{p-1} + |g|\cdot|f+g|^{p-1},\end{equation*}
and integration over XX to find that
∫X∣f(x)+g(x)∣p dμ(x)≤∫X∣f(x)∣⋅∣f(x)+g(x)∣p−1 dμ(x)+∫X∣g(x)∣⋅∣f(x)+g(x)∣p−1 dμ(x)≤Ho¨lder’s(∥f∥Lp(X,μ)+∥g∥Lp(X,μ))⋅∥∣f+g∣p−1∥Lq(X,μ).\begin{align*}\int_X |f(x) + g(x)|^p \, \text{d} \mu(x) & \leq \int_X |f(x)| \cdot |f(x) + g(x)|^{p-1} \, \text{d} \mu(x) + \int_X |g(x)| \cdot |f(x) + g(x)|^{p-1} \, \text{d} \mu(x) \\\overset{\text{Hölder's}}{ \leq}& \left( \| f \|_{L_p(X, \mu)} + \| g \|_{L_p(X, \mu)}\right) \cdot \left\| | f +g |^{p-1}\right\|_{L^q(X, \mu)}.\end{align*}
Since (p−1)q=p(p-1)q = p,
∥∣f+g∣p−1∥Lq(X,μ)=(∫X∣f(x)+g(x)∣pdμ(x))1q\begin{equation*}\left\| | f + g |^{p-1}\right\|_{L^q(X, \mu)} = \left( \int_X | f(x) + g(x) |^{p} \text{d} \mu(x) \right)^{\frac{1}{q}}\end{equation*}
from which we obtain by dividing previous inequality by (∫X∣f+g∣p dμ)1/q\left( \int_X |f + g|^p \, d\mu \right)^{1/q}
(∫X∣f(x)+g(x)∣p dμ(x))1−1q≤∥f∥Lp(X,μ)+∥g∥Lp(X,μ)\begin{equation*}\left( \int_X |f(x) + g(x)|^p \, \text{d} \mu(x) \right)^{1 - \frac{1}{q}} \leq \| f \|_{L_p(X, \mu)} + \| g \|_{L_p(X, \mu)}\end{equation*}
Observing that 1−1/q=1/p 1 - 1/q = 1/p, we obtain ∥f+g∥p≤∥f∥p+∥g∥p \|f + g\|_p \leq \|f\|_p + \|g\|_p . □\Box
\end{proof}
\begin{corollary}
For 1≤p≤∞1 \leq p \leq \infty, Lp(X,μ)L_p(X, \mu) is a normed linear space
\end{corollary}
\begin{proof}
\quad Proof. The Minkowski's inequality means that, whenever 1≤p<∞1 \leq p < \infty, the formula
d(f,g):=∥f−g∥p\begin{equation*}d(f, g) := \|f - g\|_p\end{equation*}
defines a metric on Lp(μ)L_p(\mu), since the triangle inequality holds true
∥g−h∥p=∥g−f+f−h∥p≤∥g−f∥p+∥f−h∥p.□\begin{equation*}\|g - h\|_p = \|g - f + f - h\|_p \leq \|g - f\|_p + \|f - h\|_p. \Box\end{equation*}
\end{proof}

Completeness of Lp(X,μ)L_p(X,\mu)

This is 2.16 Theorem from [Adams2003]

We want to show that the spaces Lp(μ)L_p(\mu), where 1≤p≤∞1 \leq p \leq \infty are Banach spaces. In order to do it we need to prove that these spaces are complete, e.g. for every Cauchy sequences {fn}\left\{f_n\right\}
∀ε>0,  ∃N such that ∀n,m≥N:∥fn−fm∥p=(∫X∣fm−fn∣p dμ)1p<ε,\begin{equation*}\forall \varepsilon > 0, \; \exists N \text{ such that } \forall n, m \geq N: \|f_n - f_m\|_p = \left( \int_X | f_m - f_n |^p \, d\mu \right)^{\frac{1}{p}} < \varepsilon ,\end{equation*}
exist a limit function inside Lp(μ)L_p(\mu).
Theorem
The space Lp(X,μ)L_p(X,\mu), where 1≤p≤∞1 \leq p \leq \infty, is complete with respect to the metric ∥f−g∥p\|f - g\|_p.
\end{theorem}
Let's prove this theorem.

The Cauchy sequence

First assume 1≤p<∞1 \leq p<\infty and let {fn}\left\{f_n\right\} be a Cauchy sequence in Lp(μ)L_p(\mu). We can extract subsequence {fnj}\left\{f_{n_j}\right\} of {fn}\left\{f_n\right\} such that
∥fnj+1−fnj∥p≤12j,j=1,2,…\begin{equation*}\| f_{n_{j+1}}-f_{n_j}\|_p \leq \frac{1}{2^j}, \quad j=1,2, \ldots\end{equation*}

The convergent monotone sequence

Let's define the sequence {gn}n=1∞\{ g_n\}^{\infty}_{n=1} for x∈Xx \in X as
gm(x)=∑j=1m∣fnj+1(x)−fnj(x)∣\begin{equation*}g_m(x) = \sum_{j=1}^m | f_{n_{j+1}}(x)-f_{n_j}(x) |\end{equation*}
The sequence is monotonically increasing,
0≤g1≤g2≤⋯≤gm≤…\begin{equation*}0 \leq g_1 \leq g_2 \leq \dots \leq g_m \leq \dots\end{equation*}
and due to Minkowski's inequality for any mm
∫Xgmp(x)dμ(x)≤(∥f1∥Lp(X,μ)+∑j=1m∥fnj+1−fnj∥Lp(X,μ))p≤1,\begin{equation*}\int_X g^p_m(x) \text{d} \mu(x) \leq \left( \| f_1 \|_{L_p(X, \mu)} + \sum_{j=1}^m \| f_{n_{j+1}}-f_{n_j} \|_{L_p(X, \mu)} \right)^{p} \leq 1 ,\end{equation*}
we get that {gn}\{ g_n\} is bounded in Lp(X,μ)L_p(X, \mu). Putting gp(x)=lim⁡m→∞gmp(x)g^p(x) = \lim_{m \rightarrow \infty} g^p_m(x), we obtain by the Monotone Convergence Theorem
∫Xgp(x)dμ(x)=lim⁡m→∞∫Xgmp(x)dμ(x)≤1\begin{equation*}\int_X g^p(x) \text{d} \mu(x) = \lim_{m \rightarrow \infty} \int_X g^p_m(x) \text{d} \mu(x) \leq 1\end{equation*}
Hence gp(x)<∞g^p(x) < \infty μ−\mu-a.e on XX and g∈Lp(X,μ) g \in L_p(X, \mu).

Pointwise convergence of {fnj}\{ f_{n_j}\}

For all k≥1,k \geq 1,
∣fnm+k−fnm∣=∣fnm+k−fnm+k−1+fnm+k−1−⋯+fnm+1−fnm∣≤∑i=m+1k∣fni−fni−1∣=gm+k−gm→0μ−a.e.\begin{align*}| f_{n_{m+k}}-f_{n_m} | &= | f_{n_{m+k}} - f_{n_{m+k-1}} + f_{n_{m+k-1}} - \dots + f_{n_{m+1}}- f_{n_m} | \\& \leq \sum_{i=m+1}^{k} | f_{n_{i}}-f_{n_{i-1}} | = g_{m+k} - g_m \rightarrow 0 \quad \mu-\text{a.e.}\end{align*}
Therefore, {fnm(x)}m=1∞ \{ f_{n_{m}}(x) \}_{m = 1}^{\infty} is Cauchy sequence on real line, so it converges. Now we can define the limit
f(x)=lim⁡m→∞fnm(x)μ−a.e. on X\begin{equation*}f(x)=\lim_{m \rightarrow \infty} f_{n_m} (x) \quad \mu-\text{a.e. on } X\end{equation*}
Since ∣fnm∣≤gm≤g|f_{n_m}| \leq g_m \leq g μ−\mu-a.e, ∣f∣≤g|f| \leq g and ∣fnm−f∣≤2pgp|f_{n_m} - f| \leq 2^p g^p. Then by Dominated Convergence Theorem,
lim⁡m→∞∫X∣f−fnm∣pdμ(x)=∫Xlim⁡m→∞∣f−fnm∣pdμ(x)=0\begin{equation*}\lim_{m \rightarrow \infty} \int_{X} | f - f_{n_m} |^p \text{d} \mu(x) = \int_{X} \lim_{m \rightarrow \infty} | f - f_{n_m} |^p \text{d} \mu(x) = 0\end{equation*}

Convergence to ff

For any ϵ>0\epsilon>0 there exists NN such that if m,n≥Nm, n \geq N, then ∥fm−fn∥p<ϵ\left\|f_m-f_n\right\|_p<\epsilon. Hence, by Fatou's lemma
∫X∣f(x)−fn(x)∣pdμ(x)=∫Xlim⁡j→∞∣fnj(x)−fn(x)∣pdμ(x)≤lim inf⁡j→∞∫X∣fnj(x)−fn(x)∣pdμ(x)≤ϵp.\begin{align*}\int_{X}\left|f(x)-f_n(x)\right|^p \text{d} \mu(x) & =\int_{X} \lim _{j \rightarrow \infty}\left|f_{n_j}(x)-f_n(x)\right|^p \text{d} \mu(x) \\& \leq \liminf _{j \rightarrow \infty} \int_{X}\left|f_{n_j}(x)-f_n(x)\right|^p \text{d} \mu(x) \leq \epsilon^p.\end{align*}
Thus ∥f−fn∥p→0\left\|f-f_n\right\|_p \rightarrow 0 as n→∞n \rightarrow \infty. Therefore Lp(Ω)L_p(\Omega) is complete and so is a Banach space.

Completness L∞(X,μ)L^{\infty}(X, \mu)

Finally, if {fn}\left\{f_n\right\} is a Cauchy sequence in L∞(X,μ)L^{\infty}(X, \mu), then there exists a set A⊂XA \subset X having measure zero such that if x∉Ax \notin A, then for every n,m=1,2,…n, m=1,2, \ldots
∣fn(x)∣≤∥fn∥∞,∣fn(x)−fm(x)∣≤∥fn−fm∥∞.\begin{equation*}\left|f_n(x)\right| \leq\left\|f_n\right\|_{\infty}, \quad\left|f_n(x)-f_m(x)\right| \leq\left\|f_n-f_m\right\|_{\infty} .\end{equation*}
Therefore, {fn}\left\{f_n\right\} converges uniformly on X/AX / A to a bounded function ff. Setting f=0f=0 for x∈Ax \in A, we have u∈L∞(Ω)u \in L^{\infty}(\Omega) and ∥fn−f∥∞→0\left\|f_n-f\right\|_{\infty} \rightarrow 0 as n→∞n \rightarrow \infty. Thus L∞(X,μ)L^{\infty}(X, \mu) is also complete and a Banach space. □\Box

Corollaries

\begin{corollary}
If 1≤p≤∞1 \leq p \leq \infty, each Cauchy sequence in Lp(X,μ)L_p(X, \mu) has a subsequence converging pointwise almost everywhere on Ω\Omega.
\end{corollary}
\begin{corollary}
L2(X,μ)L^2(X, \mu) is a Hilbert space(we proved completeness) with respect to the inner product
(u,v)=∫Xu(x)v(x)‾dx.\begin{equation*}(u, v)=\int_{X} u(x) \overline{v(x)} d x .\end{equation*}
Hölder's inequality for L2(X,μ)L^2(X, \mu) is just the well-known Schwarz inequality
∣(u,v)∣≤∥u∥2∥v∥2.\begin{equation*}|(u, v)| \leq\|u\|_2\|v\|_2 .\end{equation*}
\end{corollary}

Examples

Example
Let X=[0,1]X = [0,1] and μ\mu is Lebesgue measure, and consider the subintervals
[0,12],[12,1],[0,13],[13,23],[23,1],[0,14],…\begin{equation*}\left[0, \frac{1}{2}\right], \left[\frac{1}{2}, 1\right], \left[0, \frac{1}{3}\right], \left[\frac{1}{3}, \frac{2}{3}\right], \left[\frac{2}{3}, 1\right], \left[0, \frac{1}{4}\right], \dots\end{equation*}
Let fnf_n denote the indicator function of the nn-th interval of the above sequence. Then
∥fn∥Lp→0,\begin{equation*}\|f_n\|_{L_p} \rightarrow 0,\end{equation*}
but fn(x)f_n(x) does not converge for any x∈[0,1]x \in [0,1].
Example
Set X=RX = \mathbb{R} and μ\mu is Lebesgue measure, for n∈Nn \in \mathbb{N}, set fn=1[n,n+1]f_n = 1_{[n, n+1]}. Then fn(x)→0f_n(x) \rightarrow 0 as n→∞n \rightarrow \infty , but ∥fn∥Lp=1 for p∈[1,∞)\|f_n\|_{L_p} = 1 \text{ for } p \in [1, \infty), thus, fn→0 f_n \rightarrow 0 pointwise, but not in LpL_p .
Example
Set X=[0,1]X = [0,1] and μ\mu is Lebesgue measure, for n∈Nn \in \mathbb{N} , set fn=n1[0,1n] f_n = n 1_{[0, \frac{1}{n}]} . Then fn(x)→0 f_n(x) \rightarrow 0 almost everywhere as n→∞ n \rightarrow \infty, but ∥fn∥L1=1\|f_n\|_{L^1} = 1, thus, fn→0 f_n \rightarrow 0 pointwise, but not in L1 L^1 .

Imbedding Theorem for Lp(Ω)L_p(\Omega) Spaces

This is 2.19 Theorem from [Adams2003]

The Imbedding Theorem gives an understanding of hierarchy of Lp(μ)L_p(\mu) spaces
Theorem (An Imbedding Theorem for LpL_p Spaces)
Suppose that vol⁡(Ω)=∫Ωdx<∞\operatorname{vol}(\Omega)=\int_{\Omega} \text{d} x<\infty and 1≤p≤q≤∞1 \leq p \leq q \leq \infty. If u∈Lq(Ω)u \in L^q(\Omega), then u∈Lp(Ω)u \in L_p(\Omega) and
∥u∥p≤(vol⁡(Ω))(1/p)−(1/q)∥u∥q.\begin{equation}\|u\|_p \leq(\operatorname{vol}(\Omega))^{(1 / p)-(1 / q)}\|u\|_q .\end{equation}
Hence
Lq(Ω)→Lp(Ω)\begin{equation}L^q(\Omega) \rightarrow L_p(\Omega)\end{equation}
If u∈L∞(Ω)u \in L^{\infty}(\Omega), then
lim⁡p→∞∥u∥p=∥u∥∞.\begin{equation}\lim _{p \rightarrow \infty}\|u\|_p=\|u\|_{\infty} .\end{equation}
Finally, if u∈Lp(Ω)u \in L_p(\Omega) for 1≤p<∞1 \leq p<\infty and if there exists a constant KK such that for all such pp
∥u∥p≤K\begin{equation}\|u\|_p \leq K\end{equation}
then u∈L∞(Ω)u \in L^{\infty}(\Omega) and
∥u∥∞≤K.\begin{equation}\|u\|_{\infty} \leq K .\end{equation}

Approximation Lp(Rn)L_p(\mathbb{R}^n) by Simple and Continuous Functions

This is 2.19 Theorem from [Adams2003]

Simple functions

Theorem
If p∈[1,∞) p \in [1, \infty) , then the set of simple functions f=∑i=1naiχEi f = \sum_{i=1}^n a_i \chi_{E_i} , where each Ei E_i is an element of the σ\sigma-algebra X \mathcal{X} and μ(Ei)<∞ \mu(E_i) < \infty , is dense in Lp(X,X,μ) L_p(X, \mathcal{X}, \mu) .
\begin{proof}
If f∈Lp f \in L_p , then f f is measurable; thus, there exists a sequence {ϕn}n=1∞ \{\phi_n\}_{n=1}^\infty of simple functions, such that ϕn→f \phi_n \rightarrow f almost everywhere with
0≤∣ϕ1∣≤∣ϕ2∣≤⋯≤∣f∣,\begin{equation*}0 \leq |\phi_1| \leq |\phi_2| \leq \cdots \leq |f|,\end{equation*}
i.e., ϕn \phi_n approximates f f from below.
For approximation sequence we get ∣ϕn−f∣p→0 |\phi_n - f|_p \rightarrow 0 μ\mu-a.e. and ∣ϕn−f∣≤2∣f∣ |\phi_n - f| \leq 2|f| if p∈L1 p \in L^1 , so by the Dominated Convergence Theorem, ∥ϕn−f∥Lp→0 \|\phi_n - f\|_{L_p} \rightarrow 0 .
\end{proof}

Continuous functions C0(Rn)C_0(\mathbb{R}^n)

Theorem
C0(Rn)C_0(\mathbb{R}^n) is dense in Lp(Rn)L_p(\mathbb{R}^n) if 1≤p<∞1 \leq p<\infty.
\begin{proof}
\quad Proof. Let u∈Lp(Rn)u \in L_p(\mathbb{R}^n) as we have seen in the previous theroem there exists a nonnegative sequence of simple functions {sn}\left\{s_n\right\} and s∈{sn}s \in \left\{s_n\right\} such that ∥u−s∥p<ϵ/2\|u-s\|_p<\epsilon / 2. Since ss is simple and p<∞p<\infty the support of ss has finite volume. We can also assume that s(x)=0s(x)=0 if x∈Ωcx \in \Omega^c. By Lusin's Theorem there exists ϕ∈C0(Rn)\phi \in C_0\left(\mathbb{R}^n\right) such that
∣ϕ(x)∣≤∥s∥∞ for all x∈Rn,\begin{equation*}|\phi(x)| \leq\|s\|_{\infty} \quad \text { for all } x \in \mathbb{R}^n,\end{equation*}
and
vol⁡({x∈Rn:ϕ(x)≠s(x)})<(ϵ4∥s∥∞)p.\begin{equation*}\operatorname{vol}\left(\left\{x \in \mathbb{R}^n: \phi(x) \neq s(x)\right\}\right)<\left(\frac{\epsilon}{4\|s\|_{\infty}}\right)^p .\end{equation*}
By An Imbedding Theorem
∥s−ϕ∥p≤∥s−ϕ∥∞(vol⁡({x∈Rn:ϕ(x)≠s(x)}))1/p<2∥s∥∞(ϵ4∥s∥∞)=ϵ2.\begin{align*}\|s-\phi\|_p & \leq\|s-\phi\|_{\infty}\left(\operatorname{vol}\left(\left\{x \in \mathbb{R}^n: \phi(x) \neq s(x)\right\}\right)\right)^{1 / p} \\& <2\|s\|_{\infty}\left(\frac{\epsilon}{4\|s\|_{\infty}}\right)=\frac{\epsilon}{2} .\end{align*}
It follows that ∥u−ϕ∥p<ϵ\|u-\phi\|_p<\epsilon. □\Box
\end{proof}

Separability of Lp(Rn)L_p(\mathbb{R}^n)

This is 2.21 Theorem from [Adams2003]

Theorem
Lp(Ω)L_p(\Omega) is separable if 1≤p<∞1 \leq p<\infty.
\begin{proof}
\quad Proof. For m=1,2,…m=1,2, \ldots let
Ωm={x∈Ω:∣x∣≤m and dist⁡(x,bdry⁡(Ω))≥1/m}.\begin{equation*}\Omega_m=\{x \in \Omega:|x| \leq m \text { and } \operatorname{dist}(x, \operatorname{bdry}(\Omega)) \geq 1 / m\} .\end{equation*}
Then Ωm\Omega_m is a compact subset of Ω\Omega. Let PP be the set of all polynomials on Rn\mathbb{R}^n having rational-complex coefficients, and let Pm={χmf:f∈P}P_m=\left\{\chi_m f: f \in P\right\} where χm\chi_m is the characteristic function of Ωm\Omega_m. PmP_m is dense in C(Ωm)C\left(\Omega_m\right). Moreover, ⋃m=1∞Pm\bigcup_{m=1}^{\infty} P_m is countable.
If u∈Lp(Ω)u \in L_p(\Omega) and ϵ>0\epsilon>0, there exists ϕ∈C0(Ω)\phi \in C_0(\Omega) such that ∥u−ϕ∥p<ϵ/2\|u-\phi\|_p<\epsilon / 2. If 1/m<dist⁡(supp⁡(ϕ),bdry⁡(Ω))1 / m<\operatorname{dist}(\operatorname{supp}(\phi), \operatorname{bdry}(\Omega)), then there exists ff in the set PmP_m such that ∥ϕ−f∥∞<(ϵ/2)(vol⁡(Ωm))−1/p\|\phi-f\|_{\infty}<(\epsilon / 2)\left(\operatorname{vol}\left(\Omega_m\right)\right)^{-1 / p}. It follows that
∥ϕ−f∥p≤∥ϕ−f∥∞(vol⁡(Ωm))1/p<ϵ/2\begin{equation*}\|\phi-f\|_p \leq\|\phi-f\|_{\infty}\left(\operatorname{vol}\left(\Omega_m\right)\right)^{1 / p}<\epsilon / 2\end{equation*}
and so ∥u−f∥p<ϵ\|u-f\|_p<\epsilon. Thus the countable set ⋃m=1∞Pm\bigcup_{m=1}^{\infty} P_m is dense in Lp(Ω)L_p(\Omega) and Lp(Ω)L_p(\Omega) is separable. □\Box
\end{proof}

References